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(D) Given: Efficiency $\eta = 90 \% = 0.9$,Primary voltage $V_P = 200 \ V$,Primary power $P_P = 3 \ kW = 3000 \ W$,Secondary current $I_S = 6 \ A$.Fir

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$450 \ V, 15 \ A$

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(D) Given: Efficiency $\eta = 90 \% = 0.9$,Primary voltage $V_P = 200 \ V$,Primary power $P_P = 3 \ kW = 3000 \ W$,Secondary current $I_S = 6 \ A$.First,calculate the primary current $I_P$ using $P_P = V_P 	imes I_P$:$I_P = \frac{P_P}{V_P} = \frac{3000 \ W}{200 \ V} = 15 \ A$.Next,calculate the output power (secondary power) $P_S$ using efficiency: $P_S = \eta 	imes P_P = 0.9 	imes 3000 \ W = 2700 \ W$.Finally,calculate the secondary voltage $V_S$ using $P_S = V_S 	imes I_S$:$V_S = \frac{P_S}{I_S} = \frac{2700 \ W}{6 \ A} = 450 \ V$.Thus,the secondary voltage is $450 \ V$ and the primary current is $15 \ A$.

$A$ transformer having an efficiency of $90 \%$ is working on a $200 \ V$ and $3 \ kW$ power supply. If the current in the secondary coil is $6 \ A$,the voltage across the secondary coil and the current in the primary coil respectively are:

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